Question

Care este compozitia procentuala masica si prin raport de masa a unui amestec de mono si dibromobenzen care contine 53,33% Br?

Answer 1

Notam raportul molar mono:di=x:y
MC6H5Br=72+5+80=157g/mol
MC6H4Br2=72+4+160=236g/mol
157x+236y...........100%
80x+160y..........53,33%
8000x+16000y=8373,33x+12586,66
373,33x=3413,33y => x=9y adica x:y=9:1
luam 9moli de mono si 1mol de dibromo
mamestec=9*157+236=1649g
1649g...........100%
1413g........85,68% monobromobenzen
236g..........14,32% dibromobenzen
mmono:mdibromo=1413/236=5,98